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Difference between revisions of "1990 AIME Problems/Problem 11"

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== Generalization: ==
 
== Generalization: ==
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Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = (a+1)! – 1
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Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = '''(a+1)! – 1'''
  
 
For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039
 
For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039

Revision as of 21:40, 28 October 2009

Problem

Someone observed that $6! = 8 \cdot 9 \cdot 10$. Find the largest positive integer $n^{}_{}$ for which $n^{}_{}!$ can be expressed as the product of $n - 3_{}^{}$ consecutive positive integers.

Solution 1

The product of $n - 3$ consecutive integers can be written as $\frac{(n - 3 + a)!}{a!}$ for some integer $a$. Thus, $n! = \frac{(n - 3 + a)!}{a!}$, from which it becomes evident that $a \ge 3$. Since $(n - 3 + a)! > n!$, we can rewrite this as $\frac{n!(n+1)(n+2) \ldots (n-3+a)}{a!} = n! \Longrightarrow (n+1)(n+2) \ldots (n-3+a) = a!$. For $a = 4$, we get $n + 1 = 4!$ so $n = 23$. For greater values of $a$, we need to find the product of $a-3$ consecutive integers that equals $a!$. $n$ can be approximated as $^{a-3}\sqrt{a!}$, which decreases as $a$ increases. Thus, $n = 23$ is the greatest possible value to satisfy the given conditions.

Solution 2

Let the largest of the (n-3) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-3) consecutive positive integers will be less than n!.

Key observation: Now for n to be maximum the smallest number (or starting number) of the (n-3) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-3) consecutive positive integers are (5, 6, 7…, n+1)

So we have (n+1)! /4! = n! => n+1 = 24 => n = 23

Generalization:

Largest positive integer n for which n! can be expressed as the product of (n-a) consecutive positive integers = (a+1)! – 1

For ex. largest n such that product of (n-6) consecutive positive integers is equal to n! is 7!-1 = 5039

Proof: Reasoning the same way as above, let the largest of the (n-a) consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of (n-a) consecutive positive integers will be less than n!.

Now, observe that for n to be maximum the smallest number (or starting number) of the (n-a) consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is (n+1).

So the (n-a) consecutive positive integers are (a+2, a+3, … n+1)

So we have (n+1)! / (a+1)! = n! => n+1 = (a+1)! => n = (a+1)! -1

See also

1990 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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